ITT we ask easy riddles/questions that people screw up

[devonin]

Re: ITT we ask easy riddles/questions that people screw up

Monty Hall has always always bugged me.

It doesn't matter how many doors there are, if you're shown what is behind doors 1 through X-2, no matter what you've picked, car or goat, there is no cirsumatance where they -can't- show you every door but two having goats.

If after the first door is revealed and shows a goat, you're suggesting my odds improve from 1/3 to 2/3 if I switch doors, but someone walking up is presented with just the two doors, their odds are 1/2

It's never sat well with me that you can have a 66% chance to win a car by picking a door, and I have only a 50% chance to win a car by picking the same door, simply becuase you know some other door had a goat behind it.

[devonin]

Re: ITT we ask easy riddles/questions that people screw up

Also, to this thread:

[Reincarnate]

Re: ITT we ask easy riddles/questions that people screw up

Monty Hall has always always bugged me.

It doesn't matter how many doors there are, if you're shown what is behind doors 1 through X-2, no matter what you've picked, car or goat, there is no cirsumatance where they -can't- show you every door but two having goats.

If after the first door is revealed and shows a goat, you're suggesting my odds improve from 1/3 to 2/3 if I switch doors, but someone walking up is presented with just the two doors, their odds are 1/2

It's never sat well with me that you can have a 66% chance to win a car by picking a door, and I have only a 50% chance to win a car by picking the same door, simply becuase you know some other door had a goat behind it.

It's because the selection space has changed as a result of having additional information.

If I am just choosing between two doors blindly, then of course it's 50/50.

The three-door scenario with the goat-door reveal is different because you started out with 3 doors -- and then, *conditional on which door you choose*, information is given to you. That information changes the probabilities in a *conditional* manner that results in a different situation than simply starting out with two doors and no information. In a two-door situation, we know there is going to be one goat and one car. In a three-door one-reveal situation, the door unopened doors in question can have either two goats OR a goat and a car -- and this depends entirely on which door you choose in the first place.

Again, consider the 1-billion door scenario. After having the goat-doors revealed to you, it should be easy to intuit that the probability is not 50/50 between your door and the only remaining closed door. The probability that the other door has the car is nearly 100% because of the information given to you resultant of your choice.

[UserNameGoesHere]

Re: ITT we ask easy riddles/questions that people screw up

So how would the Monty Hall problem change if instead there were two contestants, who each choose a separate door, and the host opens the third door to reveal a goat? Would both contestants be better off in switching their answers? Even though one of them would be guaranteed a goat?

[Reincarnate]

Re: ITT we ask easy riddles/questions that people screw up

So how would the Monty Hall problem change if instead there were two contestants, who each choose a separate door, and the host opens the third door to reveal a goat? Would both contestants be better off in switching their answers? Even though one of them would be guaranteed a goat?

This problem isn't well-formed because it is possible for each contestant to select a separate goat, and therefore the remaining unopened door would have the car -- making it impossible for the host to open any unchosen goat doors.

[Emithith]

Re: ITT we ask easy riddles/questions that people screw up

The problem would work better with 4 doors..

[UserNameGoesHere]

Re: ITT we ask easy riddles/questions that people screw up

This problem isn't well-formed because it is possible for each contestant to select a separate goat, and therefore the remaining unopened door would have the car -- making it impossible for the host to open any unchosen goat doors.

Let me rephrase then. There are three doors, one of which has a car, two of which have goats. Each contestant picks a separate door. Unknown to the two contestants, one of them picked the car door. The host opens the remaining door to reveal a goat. Should both contestants switch doors, regardless of the fact one would be guaranteed the goat? Assume if both ended up with the car door in the end, that a car would be provided for each.

[Reincarnate]

Re: ITT we ask easy riddles/questions that people screw up

Let me rephrase then. There are three doors, one of which has a car, two of which have goats. Each contestant picks a separate door. Unknown to the two contestants, one of them picked the car door. The host opens the remaining door to reveal a goat. Should both contestants switch doors, regardless of the fact one would be guaranteed the goat? Assume if both ended up with the car door in the end, that a car would be provided for each.

The situations of me having the car vs. you having the car are isomorphic to each other. If we are confining this situation to one where either I have the car or you have the car, we're removing a lot of the Bayesian weirdness because we aren't given any new information based on choices. The end situation is no different from one where we are given two doors and we're told that one has a goat and the other has a car, where we must choose blindly.

At that point, it's all equal probability. You can switch or stay if you want to -- it won't matter.

[LongGone]

Re: ITT we ask easy riddles/questions that people screw up

If after the first door is revealed and shows a goat, you're suggesting my odds improve from 1/3 to 2/3 if I switch doors, but someone walking up is presented with just the two doors, their odds are 1/2

That someone walking up and seeing two doors are not looking at equal probability doors, its already rigged to begin with due to the previous condition.

If you came into an room and see two closed doors A and B then I tell you that there is a car behind one, without any further information your chance of opening the door with the car is of course 50/50. This is because you'd randomly pick either door. To -you-, both doors are equal.

However if from that point I tell you -additional- information "hey you know door A has a 99% chance of having a car I swear to god", you do not have to rely on random picking anymore; there is now a definite difference between door A and B. After hearing what I said, if you didn't believe me and still randomly picked a door its still 50/50 of course.

[1] P(I pick Door A, Car behind it)= 1/2 * 99/100
[2] P(I pick Door A, Goat behind it)=1/2 * 1/100
[3] P(I pick Door B, Car behind it)=1/2 * 1/100
[4] P(I pick Door B, Goat behind it)=1/2 * 99/100

The probability I get a car is case [1]+[3], which is 1/2. Likewise the probability I get a goat is case [2]+[4], which is also 1/2.

Now if you trusted my information (which is correct), you would pick only door A. Which means, the first stage of the random probability (picking a door) is out the window, because you are picking door A. The only probability left is that 99% chance if the car is behind the door, or not.

[UserNameGoesHere]

Re: ITT we ask easy riddles/questions that people screw up

The situations of me having the car vs. you having the car are isomorphic to each other. If we are confining this situation to one where either I have the car or you have the car, we're removing a lot of the Bayesian weirdness because we aren't given any new information based on choices. The end situation is no different from one where we are given two doors and we're told that one has a goat and the other has a car, where we must choose blindly.

At that point, it's all equal probability. You can switch or stay if you want to -- it won't matter.

So then the mere act of observation or, rather, of even mere thought changes the outcome of a physical condition? (In this case, changes probabilities?) The only difference between the problem I stated and the initial problem is the addition of a second actor.

[sanzath]

Re: ITT we ask easy riddles/questions that people screw up

You walk along a yellow brick road (lmfao) to try and find a pot of gold. On your journey, you met two leprechauns standing in front of two diverging paths, one which has the pot of gold at the end of the path. One of them tells the truth all the time, and the other lies all the time. They will only let you ask one question to any one of them. What question should you ask?

Didn't see anybody answer this yet, so I may as well. You ask - "Which path would the other leprechaun tell me to take?" If you're talking to the truth-chaun, he'll tell you the other one's lie, and if you're talking to the lie-chaun, he'll lie about the other one's truth. So in either case they'll be telling you to go down the wrong path, so you just pick the other one. Yay, pot of gold!

[Reincarnate]

Re: ITT we ask easy riddles/questions that people screw up

So then the mere act of observation or, rather, of even mere thought changes the outcome of a physical condition? (In this case, changes probabilities?) The only difference between the problem I stated and the initial problem is the addition of a second actor.

The mere act of conditionally-achieved information is what matters.

The difference is *not* merely adding a second actor. The main difference is the fact that your new setup removes the conditional by forcing at least one of the two actors to choose a door with a car behind it. By the very way you've defined the problem, there is no conditional probability present. The situation will ALWAYS end up the same way: You vs. me with a goat behind one door and a car behind the other. As a result, the probabilities are simply 50/50.

[Ninty64]

Re: ITT we ask easy riddles/questions that people screw up

Monty Hall problem <3

It bugged me when I first learned it, but it makes sense now.

It's because of what you know. You pick 1 door. So there's a 66% chance that the goat will be behind one of the other doors. So when they show one of those doors that is incorrect, then there's still that 66% chance in the other door because of the original set-up.

just had to give my 2cents on that.

[iironiic]

Re: ITT we ask easy riddles/questions that people screw up

Didn't see anybody answer this yet, so I may as well. You ask - "Which path would the other leprechaun tell me to take?" If you're talking to the truth-chaun, he'll tell you the other one's lie, and if you're talking to the lie-chaun, he'll lie about the other one's truth. So in either case they'll be telling you to go down the wrong path, so you just pick the other one. Yay, pot of gold!

Yay!! *\d:3/*

[nois-or-e]

Re: ITT we ask easy riddles/questions that people screw up

ITT we ask easy riddles/questions that people screw up

heh